Answer:
[tex]\mu = 0.31[/tex]
Explanation:
As we know that skier start from the foot of the inclined plane with speed
v = 12 m/s
the angle of the inclined plane is
[tex]\theta = 18 degree[/tex]
length of the inclined plane = 12.2 m
final speed of the skier = 0
so here we can use energy conservation
Work done against gravity + work done against friction = change in kinetic energy
so we will have
[tex]-mgLsin\theta - \mu mg Lcos\theta = 0 - \frac{1}{2}mv^2[/tex]
[tex](9.81)(12.2)sin18 + (\mu)(9.81)(12.2)cos(18) = \frac{1}{2}(12^2)[/tex]
[tex]36.98 + \mu(113.8) = 72[/tex]
[tex]\mu = 0.31[/tex]
The average coefficient of friction is mathematically given as
u=0.31
Question Parameters:
A skier traveling 12.0 m/s reaches the foot of a steady upward 18.0º
incline and glides 12.2 m up along this slope before coming to rest.
Generally the equation for the Work done against gravity is mathematically given as
Wg = change in kinetic energy - work done against friction
Where
Wg=-mgLsin\theta
Therefore
[tex]-mgLsin\theta=0-1/2mv^2-(-mgLcos\theta)[/tex]
(9.81)(12.2)sin18= \frac{1}{2}(12^2)-( (u)(9.81)(12.2)cos(18) )
u=0.31
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